Statics, centroids of lines, areas and volumes


by jonjacson
Tags: areas, centroids, lines, statics, volumes
jonjacson
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#1
Apr30-13, 12:55 PM
P: 130
Hi to everybody.

Iīm reading a book about statics and I cannot understand this chapter. I have been calculating moments of forces in hundreds of problems, when I found a force acting on a body I needed to fix a coordinate system, then calculating the moment arms of that force around a point and then using the equation M= r x f, or M=fd if it was a simple case on a plane.

Now they start talking about calculating the center of gravity and using the moments principle (the sum of the moments has a resultant that passes throught the center of gravity) they give this equation:

x = ∫xdw/ W

The point is that x is the moment arm of the force dw (differential weight) around the x axis if I am right.

Well if that is true, What is the meaning of this advice some pages later in the book?



I see the logic, you need to use the coordinate of the centroid of the differential element when you are integrating, but I cannot understand examples 5.12 a, and 5.12 b.

In the first example the only option to see any logic for those moment arms is assuming that the gravity is acting on the z axis at point C. Until here it could be right.

But now I cannot understand in any way example 5.12 b, it says that the moment arm of the differential element has a moment arm of xc around axis x, if that is true WHat is the axis in which the gravity is acting??

There are three options:

1.- If gravity acts on the x axis to point C, we have a force parallel to the x axis so the moment is zero, this cannot be the case.

2.- If gravity acts on the y axis to point C, the moment arm is zc but the book says clearly that zc is not the moment arm, that it is xc, so this cannot be the case.

3.- So the only option should be gravity acting on the z axis to point C, but that would cross the x axis since due to symmetry around z axis.

So I understand you need to use the centroids of the differential elements, not the boundaries, but in this case I cannot see in any way that zc is the moment arm around x axis for that element.

Does anybody know the line of action of the gravity in that example? And what about the moment arms of the element of 5.12 b?

Maybe Iīm confused and it has some logic, but now I cannot understand this.
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jonjacson
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#2
May1-13, 02:36 AM
P: 130
Does anybody know something about this?
SteamKing
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#3
May1-13, 10:22 AM
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In 5.12a, pretend the gravity acts at point C perpendicular to the image (parallel to the z-axis).

In 5.12b, to calculate zc, pretend gravity is acting parallel with the x-axis perpendicular to the y-z plane. To calculate xc, pretend gravity is acting parallel with the z-axis perpendicular to the x-y plane.

jonjacson
jonjacson is offline
#4
May1-13, 04:06 PM
P: 130

Statics, centroids of lines, areas and volumes


Quote Quote by SteamKing View Post
In 5.12a, pretend the gravity acts at point C perpendicular to the image (parallel to the z-axis).

In 5.12b, to calculate zc, pretend gravity is acting parallel with the x-axis perpendicular to the y-z plane. To calculate xc, pretend gravity is acting parallel with the z-axis perpendicular to the x-y plane.
Thank you very much for your answer!!!!

Ups I see, when you calculate xc, yc and zc you imagine an orthogonal direction for the gravity, in an hypothetical case. Then in reality, gravity will act only on a particular direction and you will use only one of those coordinates, thanks.

Now just another question, I feel bad asking this because it seems too basic, in 5.12 a for example, the moment arm of the differential element around the x axis I think it is yc, not xc.

As I understand it the moment arm around an axis for a given force is the shortest distance, if we have a force on the z axis at point C, the shortest distance to the x axis is yc, not xc. Maybe I have a bad day and I donīt understand anything, I donīt know, but I cannot see how xc is used as moment arm around the x axis.

The same happens with the y axis, in that case I see that the moment arm is xc, so the equations should look like this:

[itex]\overline{x}[/itex] = ∫ yc dA / A

[itex]\overline{y}[/itex] = ∫ xc dA/ A

I donīt understand why they are using the distance yc as moment arm around the y axis, and xc for the x axis.
SteamKing
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#5
May1-13, 06:20 PM
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P: 5,549
I don't understand your comments about the moment arm being the 'shortest' distance.

The centroidal distance xc is the location along the x-axis about which the area, volume, mass, etc. is balanced. Another way to look at it is as a weighted average, where the weighting factor is the distance along the x-axis from the origin. Ditto for yc and the y-axis and zc and the z-axis.
jonjacson
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#6
May2-13, 03:12 AM
P: 130
I have just realized what was my mistake, I misread a sentence and I thought that this equation:

[itex]\bar{x}[/itex] = ∫x dW / W

Was for the moment around the x axis, thatīs why I didnīt understand the moment arms, in fact this is the equation for the moments around the y axis and now everything makes sense.

Iīm sorry for my mistakes :( .

Thank you very much, and see you in other threads!!


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